Homework Helpers: Physics

9 Heat and Thermodynamics

Answer Key

The actual answers will be shown in brackets, followed by an explanation. If you don’t understand an explanation that is given in this section, you may want to go back and review the lesson that the question came from.

Lesson 9–1 Review

1. [Temperature]—Remember, the keyword in this def, inition is average.

2. [3.0 atm]—Because our solution to this problem will not involve one of the constants, we can use the units that came with the problem, with the exception for the temperature, which must always be converted to kelvin.

Convert: T1 = 24.0°C = (24.0 + 273) = 297 K

T2 = 49.0°C = (49.0 + 273) = 332 K

Given: V1 = 6.65 cm3 T1 = 297 K P1 = 2.0 atm V2 = 5.00 cm3 T2 = 332 K

Find: P2

3. [9.92 × 1025]—When you want to solve for the number of particles in a sample, we use the formula PV = NkB T, which contains the Boltzmann constant (kB), with a value of 1.38 × 10–23 J/K. Remember: As with our other gas law constant, keep in mind that 1 J = (1 × 10–3 m3) · Pa. This means that we can use the units for volume and pressure that came with our problem. We only need to convert the temperature to kelvin.

4. [a. absolute zero]—Although this temperature has never been achieved, it can be calculated.

5. [h. pressure]—Pressure is often measured in N/m2.

6. [b. The pressure would increase by a factor of three.]—According to the ideal gas law, the pressure and temperature of the gas will vary directly, so tripling the temperature at constant volume will triple the pressure.